Sum of maximum of all subarrays by adding even frequent maximum twice
Last Updated :
23 Jul, 2025
Given an array arr[] consisting of N integers (All array elements are a perfect power of 2), the task is to calculate the sum of the maximum elements in all the subarrays.
Note: If the frequency of the maximum element in a subarray is even, add twice the value of that element to the sum.
Examples:
Input: arr[] = {1, 2}
Output: 5
Explanation: All possible subarrays are {1}, {1, 2}, {2}.
Subarray 1: {1}. Maximum = 1. Sum = 1.
Subarray 2: {1, 2}. Maximum = 2. Sum = 3.
Subarray 3: {2}. Maximum = 2.Sum = 5.
Therefore, required output is 5.
Input: arr[] = {4, 4}
Output: 16
Explanation: All possible subarrays are {4}, {4, 4}, {4}.
Subarray 1: {4}. Maximum = 4. Sum = 1.
Subarray 2: {4, 4}. Maximum = 4. Since the maximum occurs twice in the subarray, Sum = 4 + 8 = 12.
Subarray 3: {4}. Maximum = 4. Sum = 16.
Therefore, required output is 16.
Naive Approach: The simplest approach to solve this problem is to generate all possible subarrays of the given array and find the maximum element in all subarrays along with the count of their occurrences. Finally, print the sum of all the maximum elements obtained. Follow the steps below to solve the problem:
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to calculate sum of
// maximum of all subarrays
void findSum(vector<int>a)
{
// Stores the sum of maximums
int ans = 0;
// Traverse the array
for(int low = 0;
low < a.size();
low++)
{
for(int high = low;
high < a.size();
high++)
{
// Store the frequency of the
// maximum element in subarray
int count = 0;
int maxNumber = 0;
// Finding maximum
for(int i = low;
i <= high; i++)
{
// Increment frequency by 1
if (a[i] == maxNumber)
count++;
// If new maximum is obtained
else if (a[i] > maxNumber)
{
maxNumber = a[i];
count = 1;
}
}
// If frequency of maximum
// is even, then add 2*maxNumber.
// Otherwise, add maxNumber
ans += maxNumber * ((count % 2 == 0) ? 2 : 1);
}
}
// Print the sum obtained
cout << (ans);
}
// Driver Code
int main()
{
vector<int>arr = { 2, 1, 4, 4, 2 };
// Function Call
findSum(arr);
}
// This code is contributed by amreshkumar3
// Java program for the above approach
import java.io.*;
class GFG {
// Function to calculate sum of
// maximum of all subarrays
public static void findSum(int a[])
{
// Stores the sum of maximums
int ans = 0;
// Traverse the array
for (int low = 0;
low < a.length; low++) {
for (int high = low;
high < a.length;
high++) {
// Store the frequency of the
// maximum element in subarray
int count = 0;
int maxNumber = 0;
// Finding maximum
for (int i = low;
i <= high; i++) {
// Increment frequency by 1
if (a[i] == maxNumber)
count++;
// If new maximum is obtained
else if (a[i] > maxNumber) {
maxNumber = a[i];
count = 1;
}
}
// If frequency of maximum
// is even, then add 2*maxNumber.
// Otherwise, add maxNumber
ans += maxNumber
* ((count % 2 == 0) ? 2 : 1);
}
}
// Print the sum obtained
System.out.println(ans);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 2, 1, 4, 4, 2 };
// Function Call
findSum(arr);
}
}
# Python3 program for the above approach
# Function to calculate sum of
# maximum of all subarrays
def findSum(a):
# Stores the sum of maximums
ans = 0
# Traverse the array
for low in range(0, len(a)):
for high in range(low,len(a)):
# Store the frequency of the
# maximum element in subarray
count = 0
maxNumber = 0
# Finding maximum
for i in range(low, high + 1):
# Increment frequency by 1
if (a[i] == maxNumber):
count += 1
# If new maximum is obtained
elif (a[i] > maxNumber):
maxNumber = a[i]
count = 1
# If frequency of maximum
# is even, then add 2*maxNumber.
# Otherwise, add maxNumber
if count % 2:
ans += maxNumber
else:
ans += maxNumber * 2
# Print the sum obtained
print(ans)
# Driver Code
arr = [ 2, 1, 4, 4, 2 ]
# Function Call
findSum(arr)
# This code is contributed by rohitsingh07052
// C# program for the above approach
using System;
class GFG {
// Function to calculate sum of
// maximum of all subarrays
public static void findSum(int[] a)
{
// Stores the sum of maximums
int ans = 0;
// Traverse the array
for (int low = 0; low < a.Length; low++) {
for (int high = low; high < a.Length; high++) {
// Store the frequency of the
// maximum element in subarray
int count = 0;
int maxNumber = 0;
// Finding maximum
for (int i = low; i <= high; i++) {
// Increment frequency by 1
if (a[i] == maxNumber)
count++;
// If new maximum is obtained
else if (a[i] > maxNumber) {
maxNumber = a[i];
count = 1;
}
}
// If frequency of maximum
// is even, then add 2*maxNumber.
// Otherwise, add maxNumber
ans += maxNumber
* ((count % 2 == 0) ? 2 : 1);
}
}
// Print the sum obtained
Console.WriteLine(ans);
}
// Driver Code
public static void Main()
{
int[] arr = { 2, 1, 4, 4, 2 };
// Function Call
findSum(arr);
}
}
// This code is contributed by ukasp.
<script>
// JavaScript program for the above approach
// Function to calculate sum of
// maximum of all subarrays
function findSum(a) {
// Stores the sum of maximums
var ans = 0;
// Traverse the array
for (var low = 0; low < a.length; low++) {
for (var high = low; high < a.length; high++) {
// Store the frequency of the
// maximum element in subarray
var count = 0;
var maxNumber = 0;
// Finding maximum
for (var i = low; i <= high; i++) {
// Increment frequency by 1
if (a[i] === maxNumber) count++;
// If new maximum is obtained
else if (a[i] > maxNumber) {
maxNumber = a[i];
count = 1;
}
}
// If frequency of maximum
// is even, then add 2*maxNumber.
// Otherwise, add maxNumber
ans += maxNumber * (count % 2 === 0 ? 2 : 1);
}
}
// Print the sum obtained
document.write(ans);
}
// Driver Code
var arr = [2, 1, 4, 4, 2];
// Function Call
findSum(arr);
</script>
Time Complexity: O(N3)
Auxiliary Space: O(1)
Optimized Approach: To optimize the above approach, the idea is to store the prefix sums of every bit of array elements and find the frequency of the largest element in a subarray in O(1) computational complexity. This approach works as all the array elements are powers of 2.
Below is the implementation of the above approach:
C++
// c++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to find the maximum
// in subarray {arr[low], ..., arr[high]}
int getCountLargestNumber(
int low, int high, int i,
vector<vector<int>> prefixSums);
// Function to calculate the prefix
// sum array
vector<vector<int>> getPrefixSums(
vector<int> a);
// Function to calculate sum of
// maximum of all subarrays
void findSum(vector<int> a)
{
// Calculate prefix sum array
vector<vector<int>> prefixSums
= getPrefixSums(a);
// Store the sum of maximums
int ans = 0;
// Traverse the array
for (int low = 0;
low < a.size();
low++) {
for (int high = low;
high < a.size();
high++) {
// Store the frequency of the
// maximum element in subarray
int count = 0;
int maxNumber = 0;
// Store prefix sum of every bit
for (int i = 30; i >= 0; i--) {
// Get the frequency of the
// largest element in subarray
count = getCountLargestNumber(
low, high, i, prefixSums);
if (count > 0) {
maxNumber = (1 << i);
// If frequency of the largest
// element is even, add 2 * maxNumber
// Otherwise, add maxNumber
ans += maxNumber
* ((count % 2 == 0) ? 2 : 1);
break;
}
}
}
}
// Print the required answer
cout << ans;
}
// Function to calculate the prefix
// sum array
vector<vector<int>> getPrefixSums(
vector<int> a)
{
// Initialize prefix array
vector<vector<int>> prefix(32, vector<int>(a.size() + 1, 0));
// Start traversing the array
for (int j = 0; j < a.size(); j++) {
// Update the prefix array for
// each element in the array
for (int i = 0; i <= 30; i++) {
// To check which bit is set
int mask = (1 << i);
prefix[i][j + 1] += prefix[i][j];
if ((a[j] & mask) > 0)
prefix[i][j + 1]++;
}
}
// Return prefix array
return prefix;
}
// Function to find the maximum
// in subarray {arr[low], ..., arr[high]}
int getCountLargestNumber(
int low, int high, int i,
vector<vector<int>> prefixSums)
{
return prefixSums[i][high + 1]
- prefixSums[i][low];
}
// Driver Code
int main()
{
vector<int> arr = { 2, 1, 4, 4, 2 };
// Function Call
findSum(arr);
return 0;
}
// This code is contributed
// by Shubham Singh
// Java program for the above approach
import java.io.*;
class GFG {
// Function to calculate sum of
// maximum of all subarrays
public static void findSum(int a[])
{
// Calculate prefix sum array
int[][] prefixSums
= getPrefixSums(a);
// Store the sum of maximums
int ans = 0;
// Traverse the array
for (int low = 0;
low < a.length;
low++) {
for (int high = low;
high < a.length;
high++) {
// Store the frequency of the
// maximum element in subarray
int count = 0;
int maxNumber = 0;
// Store prefix sum of every bit
for (int i = 30; i >= 0; i--) {
// Get the frequency of the
// largest element in subarray
count = getCountLargestNumber(
low, high, i, prefixSums);
if (count > 0) {
maxNumber = (1 << i);
// If frequency of the largest
// element is even, add 2 * maxNumber
// Otherwise, add maxNumber
ans += maxNumber
* ((count % 2 == 0) ? 2 : 1);
break;
}
}
}
}
// Print the required answer
System.out.println(ans);
}
// Function to calculate the prefix
// sum array
public static int[][] getPrefixSums(
int[] a)
{
// Initialize prefix array
int[][] prefix = new int[32][a.length + 1];
// Start traversing the array
for (int j = 0; j < a.length; j++) {
// Update the prefix array for
// each element in the array
for (int i = 0; i <= 30; i++) {
// To check which bit is set
int mask = (1 << i);
prefix[i][j + 1] += prefix[i][j];
if ((a[j] & mask) > 0)
prefix[i][j + 1]++;
}
}
// Return prefix array
return prefix;
}
// Function to find the maximum
// in subarray {arr[low], ..., arr[high]}
public static int
getCountLargestNumber(
int low, int high, int i,
int[][] prefixSums)
{
return prefixSums[i][high + 1]
- prefixSums[i][low];
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 2, 1, 4, 4, 2 };
// Function Call
findSum(arr);
}
}
# Python program for the above approach
# Function to calculate sum of
# maximum of all subarrays
def findSum(a):
# Calculate prefix sum array
prefixSums = getPrefixSums(a);
# Store the sum of maximums
ans = 1;
# Traverse the array
for low in range(len(a)):
for high in range(len(a)):
# Store the frequency of the
# maximum element in subarray
count = 0;
maxNumber = 0;
# Store prefix sum of every bit
for i in range(30,0,-1):
# Get the frequency of the
# largest element in subarray
count = getCountLargestNumber(low, high, i, prefixSums);
if (count > 0):
maxNumber = (1 << i);
# If frequency of the largest
# element is even, add 2 * maxNumber
# Otherwise, add maxNumber
if(count % 2 == 0):
ans += maxNumber * 2;
else:
ans += maxNumber * 1;
break;
# Print the required answer
print(ans);
# Function to calculate the prefix
# sum array
def getPrefixSums(a):
# Initialize prefix array
prefix = [[0 for i in range(len(a)+1)] for j in range(32)];
# Start traversing the array
for j in range(len(a)):
# Update the prefix array for
# each element in the array
for i in range(31):
# To check which bit is set
mask = (1 << i);
prefix[i][j + 1] += prefix[i][j];
if ((a[j] & mask) > 0):
prefix[i][j + 1]+=1;
# Return prefix array
return prefix;
# Function to find the maximum
# in subarray:arr[low], ..., arr[high]
def getCountLargestNumber(low, high, i, prefixSums):
return prefixSums[i][high + 1] - prefixSums[i][low];
# Driver Code
if __name__ == '__main__':
arr = [ 2, 1, 4, 4, 2 ];
# Function Call
findSum(arr);
# This code is contributed by gauravrajput1
// C# program for the above approach
using System;
public class GFG {
// Function to calculate sum of
// maximum of all subarrays
public static void findSum(int []a) {
// Calculate prefix sum array
int[,] prefixSums = getPrefixSums(a);
// Store the sum of maximums
int ans = 0;
// Traverse the array
for (int low = 0; low < a.Length; low++) {
for (int high = low; high < a.Length; high++) {
// Store the frequency of the
// maximum element in subarray
int count = 0;
int maxNumber = 0;
// Store prefix sum of every bit
for (int i = 30; i >= 0; i--) {
// Get the frequency of the
// largest element in subarray
count = getCountLargestNumber(low, high, i, prefixSums);
if (count > 0) {
maxNumber = (1 << i);
// If frequency of the largest
// element is even, add 2 * maxNumber
// Otherwise, add maxNumber
ans += maxNumber * ((count % 2 == 0) ? 2 : 1);
break;
}
}
}
}
// Print the required answer
Console.WriteLine(ans);
}
// Function to calculate the prefix
// sum array
public static int[,] getPrefixSums(int[] a) {
// Initialize prefix array
int[,] prefix = new int[32,a.Length + 1];
// Start traversing the array
for (int j = 0; j < a.Length; j++) {
// Update the prefix array for
// each element in the array
for (int i = 0; i <= 30; i++) {
// To check which bit is set
int mask = (1 << i);
prefix[i, j + 1] += prefix[i,j];
if ((a[j] & mask) > 0)
prefix[i, j + 1]++;
}
}
// Return prefix array
return prefix;
}
// Function to find the maximum
// in subarray {arr[low], ..., arr[high]}
public static int getCountLargestNumber(int low, int high, int i, int[,] prefixSums) {
return prefixSums[i,high + 1] - prefixSums[i,low];
}
// Driver Code
public static void Main(String[] args) {
int[] arr = { 2, 1, 4, 4, 2 };
// Function Call
findSum(arr);
}
}
// This code is contributed by gauravrajput1
<script>
// javascript program for the above approach
// Function to calculate sum of
// maximum of all subarrays
function findSum(a)
{
// Calculate prefix sum array
var prefixSums = getPrefixSums(a);
// Store the sum of maximums
var ans = 0;
// Traverse the array
for (var low = 0; low < a.length; low++) {
for (var high = low; high < a.length; high++) {
// Store the frequency of the
// maximum element in subarray
var count = 0;
var maxNumber = 0;
// Store prefix sum of every bit
for (var i = 30; i >= 0; i--) {
// Get the frequency of the
// largest element in subarray
count = getCountLargestNumber(low, high, i, prefixSums);
if (count > 0) {
maxNumber = (1 << i);
// If frequency of the largest
// element is even, add 2 * maxNumber
// Otherwise, add maxNumber
ans += maxNumber * ((count % 2 == 0) ? 2 : 1);
break;
}
}
}
}
// Print the required answer
document.write(ans);
}
// Function to calculate the prefix
// sum array
function getPrefixSums(a) {
// Initialize prefix array
var prefix = Array(32).fill().map(()=>Array(a.length + 1).fill(0));
// Start traversing the array
for (var j = 0; j < a.length; j++) {
// Update the prefix array for
// each element in the array
for (var i = 0; i <= 30; i++) {
// To check which bit is set
var mask = (1 << i);
prefix[i][j + 1] += prefix[i][j];
if ((a[j] & mask) > 0)
prefix[i][j + 1]++;
}
}
// Return prefix array
return prefix;
}
// Function to find the maximum
// in subarray {arr[low], ..., arr[high]}
function getCountLargestNumber(low , high , i, prefixSums) {
return prefixSums[i][high + 1] - prefixSums[i][low];
}
// Driver Code
var arr = [ 2, 1, 4, 4, 2 ];
// Function Call
findSum(arr);
// This code is contributed by gauravrajput1
</script>
Time Complexity: O(N2)
Auxiliary Space: O(32 * N)
Efficient Approach: To optimize the above approach, the idea is to use the property that all the array elements are powers of 2, and leverage that property to solve the problem. Follow the steps below to solve the problem:
- Iterate through all powers of 2 in descending order. Consider any arbitrary power of 2 as a mask.
- Divide the array into subarrays such that no subarray will contain arr[index] = -1, where an index is any valid position in the array.
- Let the subarray obtained from the above step be S. Traverse through S and add the values contributed by only the subarrays in S, which have the current mask, from the outer loop. Also set the corresponding position, where arr[index] = mask, to arr[index] = -1.
- To calculate the values contributed by all the subarrays in S that contains the mask and maintain three counters as oddCount, eventCount, and the frequency of mask.
- The pointer prev points to the previous index, such that arr[prev] = mask.
- At any index, where arr[index] = mask, get the count of integers between the last occurrence of the mask and the current occurrence by subtracting prev from the index. Use this count and the parity of frequency of mask, to get the values contributed by all contiguous subarrays that contain a mask, using the formula count = (index - prev) and add the count to the answer.
- If the frequency of the maximum is even or odd and if the parity is odd:
- Values contributed by all the contiguous subarrays that have the frequency of mask as odd is (count - 1)*evenCount + oddCount.
- Values contributed by all the contiguous subarrays that have the frequency of mask as even is 2*(count - 1)*oddCount + 2*evenCount.
- Otherwise, if the parity is even:
- Values contributed by all the contiguous subarrays that have the frequency of mask as odd is (count - 1)*oddCount + evenCount.
- Values contributed by all the contiguous subarrays that have the frequency of mask as even is 2*(count - 1) * evenCount + 2 * oddCount.
- Add all the corresponding values to the answer. Also add the count to evenCount if parity is even. Otherwise, add count to oddCount.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function that takes any subarray
// S and return values contributed
// by only the subarrays in S containing mask
int findSumOfValuesOfRange(vector<int> &a, int low, int high,
int mask)
{
if (low > high)
return 0;
// Stores count even, odd count of
// occurrences and maximum element
int evenCount = 0, oddCount = 0, countLargestNumber = 0;
int prev = low - 1, ans = 0;
// Traverse from low to high
for (int i = low; i <= high; i++) {
// Checking if this position
// in the array is mask
if ((mask & a[i])) {
// Mask is the largest
// number in subarray.
// Increment count by 1
countLargestNumber++;
// Store parity as 0 or 1
int parity = countLargestNumber & 1;
// Setting a[i]=-1, this
// will help in splitting
// array into subarrays
a[i] = -1;
int count = i - prev;
ans += count;
// Add values contributed
// by those subarrays that
// have an odd frequency
ans += (count - 1)
* ((parity) ? evenCount : oddCount);
ans += ((parity) ? oddCount : evenCount);
// Adding values contributed
// by those subarrays that
// have an even frequency
ans += 2 * (count - 1)
* ((parity) ? oddCount : evenCount);
ans += 2
* ((parity) ? evenCount : oddCount);
// Set the prev pointer
// to this position
prev = i;
if (parity == 1)
oddCount += count;
else
evenCount += count;
}
}
if (prev != low - 1) {
int count = high - prev;
int parity = countLargestNumber % 2;
ans += count
* ((parity) ? oddCount : evenCount);
ans += 2 * count
* ((parity) ? evenCount : oddCount);
ans *= mask;
}
// Return the readonly sum
return ans;
}
// Function to calculate sum of
// maximum of all subarrays
void findSum(vector<int> a)
{
int ans = 0;
int prev = -1;
// Iterate over the range [30, 0]
for (int i = 30; i >= 0; i--) {
int mask = (1 << i);
// Inner loop through the
// length of the array
for (int j = 0; j < a.size(); j++) {
// Divide the array such
// that no subarray will
// have any index set to -1
if (a[j] == -1) {
ans += findSumOfValuesOfRange(a, prev + 1,
j - 1, mask);
prev = j;
}
}
// Find the sum of subarray
ans += findSumOfValuesOfRange(a, prev + 1,
a.size() - 1, mask);
}
// Print the sum obtained
cout << ans << endl;
}
// Driver Code
int main()
{
vector<int> arr = { 2, 1, 4, 4, 2 };
// Function call
findSum(arr);
}
// This code is contributed by phasing17
// Java program for the above approach
import java.io.*;
class GFG {
// Function to calculate sum of
// maximum of all subarrays
public static void findSum(int a[])
{
int ans = 0;
int prev = -1;
// Iterate over the range [30, 0]
for (int i = 30; i >= 0; i--) {
int mask = (1 << i);
// Inner loop through the
// length of the array
for (int j = 0;
j < a.length; j++) {
// Divide the array such
// that no subarray will
// have any index set to -1
if (a[j] == -1) {
ans += findSumOfValuesOfRange(
a, prev + 1, j - 1, mask);
prev = j;
}
}
// Find the sum of subarray
ans += findSumOfValuesOfRange(
a, prev + 1, a.length - 1, mask);
}
// Print the sum obtained
System.out.println(ans);
}
// Function that takes any subarray
// S and return values contributed
// by only the subarrays in S containing mask
public static int findSumOfValuesOfRange(
int[] a, int low, int high, int mask)
{
if (low > high)
return 0;
// Stores count even, odd count of
// occurrences and maximum element
int evenCount = 0, oddCount = 0,
countLargestNumber = 0;
int prev = low - 1, ans = 0;
// Traverse from low to high
for (int i = low; i <= high; i++) {
// Checking if this position
// in the array is mask
if ((mask & a[i]) > 0) {
// Mask is the largest
// number in subarray.
// Increment count by 1
countLargestNumber++;
// Store parity as 0 or 1
int parity = countLargestNumber % 2;
// Setting a[i]=-1, this
// will help in splitting
// array into subarrays
a[i] = -1;
int count = i - prev;
ans += count;
// Add values contributed
// by those subarrays that
// have an odd frequency
ans += (count - 1)
* ((parity == 1) ? evenCount
: oddCount);
ans += ((parity == 1) ? oddCount
: evenCount);
// Adding values contributed
// by those subarrays that
// have an even frequency
ans += 2 * (count - 1)
* ((parity == 1) ? oddCount
: evenCount);
ans += 2
* ((parity == 1) ? evenCount
: oddCount);
// Set the prev pointer
// to this position
prev = i;
if (parity == 1)
oddCount += count;
else
evenCount += count;
}
}
if (prev != low - 1) {
int count = high - prev;
int parity = countLargestNumber % 2;
ans += count
* ((parity == 1)
? oddCount
: evenCount);
ans += 2 * count
* ((parity == 1)
? evenCount
: oddCount);
ans *= mask;
}
// Return the final sum
return ans;
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 2, 1, 4, 4, 2 };
// Function call
findSum(arr);
}
}
# Python program for the above approach
# Function to calculate sum of
# maximum of all subarrays
def findSum(a):
ans = 0
prev = -1
# Iterate over the range [30, 0]
for i in range(30,-1,-1):
mask = (1 << i)
# Inner loop through the
# length of the array
for j in range(len(a)):
# Divide the array such
# that no subarray will
# have any index set to -1
if (a[j] == -1):
ans += findSumOfValuesOfRange(a, prev + 1, j - 1, mask)
prev = j
# Find the sum of subarray
ans += findSumOfValuesOfRange(a, prev + 1, len(a) - 1, mask)
# Print the sum obtained
print(ans)
# Function that takes any subarray
# S and return values contributed
# by only the subarrays in S containing mask
def findSumOfValuesOfRange(a , low , high , mask):
if (low > high):
return 0
# Stores count even, odd count of
# occurrences and maximum element
evenCount,oddCount,countLargestNumber = 0,0,0
prev,ans = low - 1,0
# Traverse from low to high
for i in range(low,high + 1):
# Checking if this position
# in the array is mask
if ((mask & a[i]) > 0):
# Mask is the largest
# number in subarray.
# Increment count by 1
countLargestNumber += 1
# Store parity as 0 or 1
parity = countLargestNumber % 2
# Setting a[i]=-1, this
# will help in splitting
# array into subarrays
a[i] = -1
count = i - prev
ans += count
# Add values contributed
# by those subarrays that
# have an odd frequency
ans += (count - 1) * (evenCount if (parity == 1) else oddCount)
ans += (oddCount if (parity == 1) else evenCount)
# Adding values contributed
# by those subarrays that
# have an even frequency
ans += 2 * (count - 1) * (oddCount if (parity == 1) else evenCount)
ans += 2 * (evenCount if (parity == 1) else oddCount)
# Set the prev pointer
# to this position
prev = i
if (parity == 1):
oddCount += count
else:
evenCount += count
if (prev != low - 1):
count = high - prev
parity = countLargestNumber % 2
ans += count * (oddCount if (parity == 1) else evenCount)
ans += 2 * count * (evenCount if (parity == 1) else oddCount)
ans *= mask
# Return the final sum
return ans
# Driver Code
arr = [ 2, 1, 4, 4, 2 ]
# function call
findSum(arr)
# This code is contributed by shinjanpatra
// C# program for the above approach
using System;
public class GFG {
// Function to calculate sum of
// maximum of all subarrays
public static void findSum(int []a) {
int ans = 0;
int prev = -1;
// Iterate over the range [30, 0]
for (int i = 30; i >= 0; i--) {
int mask = (1 << i);
// Inner loop through the
// length of the array
for (int j = 0; j < a.Length; j++) {
// Divide the array such
// that no subarray will
// have any index set to -1
if (a[j] == -1) {
ans += findSumOfValuesOfRange(a, prev + 1, j - 1, mask);
prev = j;
}
}
// Find the sum of subarray
ans += findSumOfValuesOfRange(a, prev + 1, a.Length - 1, mask);
}
// Print the sum obtained
Console.WriteLine(ans);
}
// Function that takes any subarray
// S and return values contributed
// by only the subarrays in S containing mask
public static int findSumOfValuesOfRange(int[] a, int low,
int high, int mask)
{
if (low > high)
return 0;
// Stores count even, odd count of
// occurrences and maximum element
int evenCount = 0, oddCount = 0, countLargestNumber = 0;
int prev = low - 1, ans = 0;
// Traverse from low to high
for (int i = low; i <= high; i++) {
// Checking if this position
// in the array is mask
if ((mask & a[i]) > 0) {
// Mask is the largest
// number in subarray.
// Increment count by 1
countLargestNumber++;
// Store parity as 0 or 1
int parity = countLargestNumber % 2;
// Setting a[i]=-1, this
// will help in splitting
// array into subarrays
a[i] = -1;
int count = i - prev;
ans += count;
// Add values contributed
// by those subarrays that
// have an odd frequency
ans += (count - 1) * ((parity == 1) ? evenCount : oddCount);
ans += ((parity == 1) ? oddCount : evenCount);
// Adding values contributed
// by those subarrays that
// have an even frequency
ans += 2 * (count - 1) * ((parity == 1) ? oddCount : evenCount);
ans += 2 * ((parity == 1) ? evenCount : oddCount);
// Set the prev pointer
// to this position
prev = i;
if (parity == 1)
oddCount += count;
else
evenCount += count;
}
}
if (prev != low - 1) {
int count = high - prev;
int parity = countLargestNumber % 2;
ans += count * ((parity == 1) ? oddCount : evenCount);
ans += 2 * count * ((parity == 1) ? evenCount : oddCount);
ans *= mask;
}
// Return the readonly sum
return ans;
}
// Driver Code
public static void Main(String[] args) {
int[] arr = { 2, 1, 4, 4, 2 };
// Function call
findSum(arr);
}
}
// This code is contributed by gauravrajput1
<script>
// javascript program for the above approach
// Function to calculate sum of
// maximum of all subarrays
function findSum(a) {
var ans = 0;
var prev = -1;
// Iterate over the range [30, 0]
for (var i = 30; i >= 0; i--) {
var mask = (1 << i);
// Inner loop through the
// length of the array
for (var j = 0; j < a.length; j++) {
// Divide the array such
// that no subarray will
// have any index set to -1
if (a[j] == -1) {
ans += findSumOfValuesOfRange(a, prev + 1, j - 1, mask);
prev = j;
}
}
// Find the sum of subarray
ans += findSumOfValuesOfRange(a, prev + 1, a.length - 1, mask);
}
// Print the sum obtained
document.write(ans);
}
// Function that takes any subarray
// S and return values contributed
// by only the subarrays in S containing mask
function findSumOfValuesOfRange(a , low , high , mask) {
if (low > high)
return 0;
// Stores count even, odd count of
// occurrences and maximum element
var evenCount = 0, oddCount = 0, countLargestNumber = 0;
var prev = low - 1, ans = 0;
// Traverse from low to high
for (var i = low; i <= high; i++) {
// Checking if this position
// in the array is mask
if ((mask & a[i]) > 0) {
// Mask is the largest
// number in subarray.
// Increment count by 1
countLargestNumber++;
// Store parity as 0 or 1
var parity = countLargestNumber % 2;
// Setting a[i]=-1, this
// will help in splitting
// array into subarrays
a[i] = -1;
var count = i - prev;
ans += count;
// Add values contributed
// by those subarrays that
// have an odd frequency
ans += (count - 1) * ((parity == 1) ? evenCount : oddCount);
ans += ((parity == 1) ? oddCount : evenCount);
// Adding values contributed
// by those subarrays that
// have an even frequency
ans += 2 * (count - 1) * ((parity == 1) ? oddCount : evenCount);
ans += 2 * ((parity == 1) ? evenCount : oddCount);
// Set the prev pointer
// to this position
prev = i;
if (parity == 1)
oddCount += count;
else
evenCount += count;
}
}
if (prev != low - 1) {
var count = high - prev;
var parity = countLargestNumber % 2;
ans += count * ((parity == 1) ? oddCount : evenCount);
ans += 2 * count * ((parity == 1) ? evenCount : oddCount);
ans *= mask;
}
// Return the final sum
return ans;
}
// Driver Code
var arr = [ 2, 1, 4, 4, 2 ];
// Function call
findSum(arr);
// This code is contributed by gauravrajput1
</script>
Time Complexity: O(30*N)
Auxiliary Space: O(1)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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